Question

A billing company that collects bills or doctors' offices in the area is conce ned that the percentage of bills being paid by medical insurance has risen Historically, that percentage has been 36%. An examination o 8,942 recent, randomly selected bills reveals that 37% of these bills are being paid by medical insurance. Is there evidence of a change in the percent of bills being paid by medical insurance? (Consider a P-value around 1% to represent reasonable evidence.) Complete parts a through e below. HA: p <0.36 Ho: p-0.37 HA p 0.37 HA: p> 0.36 Ho: p-0.37 HA: p > 0.37 HA: p<0.37 H: p 0.36 HA p 0.36 b) Check the assumptions and conditions. Select all that apply. The Randomization Condition is met. The Independence Assumption is met. C. The Success Failure Condition is met. A. y B. D. The 1096 Condition is met. c) Perform the test and find the P-value. What is the test statistic? z 1.97 (Round to two decimal places as needed.) What is the P-value? P-value Round to three decimal places as needed.)

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H₀ : p = 0.36

H_a : p $\neq$ 0.36

n = 8942

$\hat p$ = 0.37

P₀ = 0.36

1 - P₀ = 0.74

Test statistic = z

= $\hat p$ - P₀ / [$\sqrt$P_{0 *} (1 - P₀ ) / n]

= 0.37 - 0.36 / [$\sqrt$(0.36 * 0.74) / 8942]

z = 1.83

P(z > 1.83) = 1 - P(z < 1.83) = 0.0336

P-value = 2 * 0.0336 = 0.0674

$\alpha$ = 0.01

P-value > $\alpha$

Fail to reject the null hypothesis .

There is not sufficient evidence